It
is quite interesting to note that any root of a polynomial with integer
coefficients and leading coefficient 1 is either an integer or
irrational. A key component of the proof is the Rational Root Theorem,
which states that if we have a polynomial with integer coefficients,
then any rational root r/s (in simplest form) has r as a factor of the
constant term and s as a factor of the leading coefficient. We can
prove it as follows. Consider the roots of a general polynomial with
integer coefficients: a _{n}x^{n} + a_{n - 1}x^{n - 1} + ... + a_{2}x^{2} + a_{1}x^{1} + a_{0} = 0If we substitute a rational root r/s (again, in simplest form) into this equation, and then multiply through by s ^{n-1}, we get:a _{n}r^{n}/s + a_{n - 1}r^{n - 1} + ... + a_{2}r^{2}s^{n-3} + a_{1}rs^{n-2} + a_{0}s^{n-1} = 0Because all the coefficients a _{i}
(for all i between 0 and n) are integers, we see right away that all
terms except the first are integers. This actually forces the first
term a_{n}r^{n}/s to be an integer as well, since everything sums to zero. For this to happen, s must divide a_{n}r^{n}; since we assume that r and s are relatively prime, the only possibility is that s divides the leading coefficient a_{n}.Now we return to the original equation, substitute r/s for x, and this time multiply through by s ^{n}/r, yielding:a _{n}r^{n-1} + a_{n - 1}r^{n - 2}s + a_{n-2}r^{n-3}s^{2} + ...+ a_{1}s^{n-1} + a_{0}s^{n}/r = 0By the same reasoning, a _{0}s^{n}/r is an integer, so r divides a_{0}s^{n}/r; again, since r and s are relatively prime, r must divide the constant term a_{0}. Combining the two parts completes the proof. (Source: The Math Forum, Ask Dr. Math, <http://mathforum.org/library/drmath/view/53226.html>, accessed 13 July 2009)Back to the problem at hand, considering a polynomial with integer coefficients and leading coefficient 1, we see that all rational roots are of the form p/1 or -p/1, by the Rational Root Theorem. Therefore, all rational roots of this polynomial are integers. The only other possibility is that of irrational roots. (If we are given a polynomial with a constant term of 0, we simply factor out the highest power of x possible, and then proceed with the proof. For example, we would rewrite the polynomial x ^{4} - 5x^{3} + 2x^{2} as x^{2}(x^{2} - 5x + 2) and then consider the x^{2} -
5x + 2 portion. The only change is that the original polynomial has a
double root of x = 0, which, however, does not interfere with the
proof as this root would simply fall into the integer category.)As an example of this occurence, we can consider the case of the polynomial x ^{3} + 2x^{2}
- 10x + 7. All possible rational roots are 1/1, 7/1, and their
opposites; we find that only x = 1 works. Factoring out (x - 1) from
the expression, we get:(x - 1)(x ^{2} + 3x - 7)and applying the quadratic formula to the second-degree portion yields: x = (-3 + √(3 ^{2} - 4 * 1 * (-7)))/(2 * 1) = (-3 + √37)/2and x = (-3 - √(3 ^{2} - 4 * 1 * (-7)))/(2 * 1) = (-3 - √37)/2. |